ANOTHER HAM SANDWICH IN THE PLANE
ALEXEY BALITSKIY, ALEXEY GARBER, AND ROMAN KARASEV
arXiv:1307.7698v3 [math.MG] 20 Dec 2013
Abstract. We show that every two nice measures in the plane can be partitioned into equal
halves by translation of an angle from arbitrary k-fan when k is odd and in some cases when k
is even. We also give some counterexamples for certain fans and measures.
1. Introduction
In [3, Theorem 1] Rade Živaljević proved that it is possible to cut a half of every one of d nice
measures in Rd by a union of several cones of a simple fan translated by some vector. A simple
fan is built of cones on facets of a simplex and a union of several cones is called a “curtain”
in [3]. Throughout this paper we call a measure µ in Rd nice if it is normalized with µ(Rd ) = 1,
has compact support, and every affine hyperplane H has µ(H) = 0.
In particular, in the two-dimensional case the curtain partition theorem can be reformulated
as follows: For every 3-fan F = {F1 , F2 , F3 } and for every two nice measures µ1 and µ2 in R2
there are index i and translation t + Fi of an angle of F such that µ1 (t + Fi ) = µ2 (t + Fi ) = 21 .
Here by an angle we mean a set consisting of a vertex, two rays from this vertex, and everything
between the rays. A particular case of an angle will be a halfplane.
In this note we generalize this theorem for the case of k-fans for odd k:
Theorem 1.1. Let k ≥ 3 be odd. For every k-fan F = {F1 , F2 , . . . , Fk } and for every two nice
measures µ1 and µ2 in R2 there are index i and a translation t + Fi of an angle of F such that
1
µ1 (t + Fi ) = µ2 (t + Fi ) = .
2
And for some symmetric fans:
Theorem 1.2. Theorem 1.1 remains true for centrally symmetric fans of 2k angles assuming
that k is even.
The proofs of these planar theorems are rather elementary by a simple continuity argument,
not requiring the Borsuk–Ulam type theorems like in [1, 2, 3]. What is more interesting, it is
possible to find counterexamples in the cases not covered by these theorems, see Section 3. In
the last Section 4 we show that it is impossible to cut half of two measures with non-compact
support with translation of an angle in general case, but some generalization of our results in
that case is still possible.
Acknowledgments. The authors thank Alfredo Hubard and the unknown referee for useful
remarks and corrections.
2. The proofs
In what follows we make the standard assumptions: The measures are taken to have continuous densities and their supports are assumed to be compact and connected. The general case
follows by a standard compactness argument, that we outline as follows. Let all supports of
1991 Mathematics Subject Classification. 52C35, 60D05.
Key words and phrases. Ham sandwich theorem.
The work of A. Garber is supported by the Russian Foundation of Basic Research grants 11-01-00633-a and
11-01-00735-a, and the Russian government project 11.G34.31.0053.
The work of R. Karasev is supported by the Dynasty foundation, the President’s of Russian Federation grant
MD-352.2012.1, and the Russian government project 11.G34.31.0053.
1
2
ALEXEY BALITSKIY, ALEXEY GARBER, AND ROMAN KARASEV
the measure be contained in a big disk D. Then for every measure µi , by a convolution with a
smooth kernel and a slight modification, we find a sequence of measures µki , having continuous
densities and connected supports in D, and converging to µi . The convergence is in the sense
that for any angle A we have µki (A) → µi (A). Now for every k we have an equipartitioning
angle Ak , and we may assume that the intersections Ak ∩D converge to an intersection A∩D in
the Hausdorff metric, where A is some angle. Assuming the contrary, without loss of generality
µi (A) < 12 , we see that for a small ε-neighborhood Aε of A the inequality µi (Aε ) < 21 retains.
Here we use that µi of the boundary ∂A and translates of this boundary slightly outside A is
zero. Now, for sufficiently large k, the convolution kernel’s support may be assumed to lie in
the ε-neighborhood of zero, and therefore we obtain µki (Aε ) < 21 . Also, for sufficiently large k,
the angle Ak is inside Aε and therefore µki (Ak ) < 21 , which is a contradiction.
Under the above assumption of continuous density and connected support there is a unique
halving line of given direction for any one of the measures. Let ej , j = 1, . . . , k be vectors
collinear to the rays of the fan F so that the angle Fj lies between rays collinear to ej and ej+1
(counter-clockwise direction and cyclic indices).
−−→
Definition 2.1. Denote Sji the set of all points M such that µi (OM + Fj ) = 21 . We will call
this set the set of j-th color for the measure µi .
i
i
Lemma 2.2. The set Sji contains two infinite rays rj,1
and rj,2
with directions of vectors −ej
i
i
i
i
and −ej+1 that are connected by a continuous curve γj in Sj . Moreover, we have rj,2
= rj+1,1
.
Proof. Let lji be the line of direction ej that divides the measure µi into equal parts and M an
arbitrary point on this line. The function
−−→
f (α) = µi (OM − αej + Fj )
of positive real α is continuous and equal to 21 for sufficiently large α because the measure µi
i
has compact support. This construction proves the existence of a ray rj,1
.
i
By the same reason we can find a suitable ray rj−1,2 on the same line lji and take joint part
i
i
of two rays as desired rj,1
= rj−1,2
.
And the last point of this lemma (existence of a curve) is also obvious since for every prescribed projection of M onto the direction perpendicular to the bisector of Fj there is a unique
such M in Sji and this M depends continuously on the projection.
i
i
Definition 2.3. Let αji be the angle between rays rj,1
and rj,2
. These rays could have different
starting points, but in that case we can take two lines containing them and choose as αji the
i
i
angle with sides that have unbounded intersections with rays rj,1
and rj,2
. This angle is obtained
from Fj by a central symmetry and a translation.
i
i
If rays rj,1
and rj,2
have opposite directions then construction of αji is obvious and it is also
obtained from Fj by a central symmetry and translation.
Proof of Theorem 1.1. Assume that for some j the angle αj1 is not a subset of the angle αj2 and
αj2 is not a subset of αj1 . Since these angles are translates of each other, this assumption means
that their boundaries intersect and either they have a whole ray of intersection, or they intersect
1
1
2
2
transversally precisely once. Then two curves rj,1
∪ rj,2
∪ γj1 and rj,1
∪ rj,2
∪ γj2 have nonempty
1
2
intersection too. Indeed, they differ from the boundaries ∂αj and ∂αj by modifications on
compact parts. If the boundaries had an infinite ray of intersections then these curves will do
the same. If the intersection was transversal and unique then these new curves will have an
odd intersection index, and therefore have nonempty intersection.
So any point
1
1
2
2
X ∈ (rj,1
∪ rj,2
∪ γj1 ) ∩ (rj,1
∪ rj,2
∪ γj2 )
can serve as the desired vertex of a translated angle Fj .
Now we may assume that for every j one of αj1 and αj2 strictly contains the other. Without
loss of generality, take some j and assume that αj1 ⊂ αj2 . Then boundaries of these angles are
ANOTHER HAM SANDWICH IN THE PLANE
3
not intersecting. Now we trace their “left” boundaries towards the infinity and note that ∂αj1
is to the “right” of ∂αj2 . Here “left” and “right” are brief substitutes of “anticlockwise” and
1
2
“clockwise”. But the “left” parts of ∂αj1 and ∂αj2 are also the “right” parts of ∂αj+1
and ∂αj+1
2
1
respectively; hence αj+1
turns out to be a subset of αj+1
.
1
2
2
1
So we have shown that if αj contains αj then αj+1 contains αj+1
and vice versa. But this
relation is impossible to extend over the cyclic order since k is odd.
Proof of Theorem 1.2. The proof follows the same lines. We also conclude that the inclusions
αj1 ⊂ αj2 and αj2 ⊂ αj1 alternate when j goes around the circle. Since k is even, we have, without
loss of generality,
1
2
α11 ⊂ α12 and αk+1
⊂ αk+1
.
(2.1)
1
1
But the angles α1 and αk+1 share the common vertex and are opposite to each other, because
they both are defined by the halving lines of µ1 in two directions. The same applies to α12 and
2
αk+1
and we conclude that (2.1) is impossible.
The above proof actually works in a slightly more general case:
Theorem 2.4. The theorem 1.2 remains true for a possibly non-symmetric fan, that contains
two opposite rays with even number of angles on one side of these rays.
Proof. The same argument as for the symmetric fans with 4k angles works for this case too.
Assume that vectors ei and ei+2m has opposite directions. Without loss of generality we can
assume that
2
1
.
(2.2)
⊂ αi+2m
αi1 ⊂ αi2 and αi+2m
1
1
But the angles αi and αi+2m share the common line and lie on opposite sides of this line. The
2
same applies to αi2 and αi+2m
, and the construction from 2.2 is impossible.
3. Examples of measures and fans
Example 3.1. We start from an example of two measures and a 4-fan shown on the figure 1.
Each red point (small disc) represents one third of the measure µ1 , they are located in the
midpoints of the regular triangle represented by dashed lines. And each blue point represents
one third of the measure µ2 , here one point lies in the center of the red triangle. One can easily
check that no angle from the fan on the right of Figure 1 can divide each measure on two equal
parts after a translation (all angles of this fan are parallel to sides of the dashed triangle on the
left part).
Indeed, for the top angle we observe that once its copy contains a blue point then it already
contains two red points in its interior. For the angle to the left (and the one to the right), we
observe that once it touches at least two red points then it already contains in its interior some
two of the blue points. For the angle at the bottom we observe that once it touches some two
of the blue points then it already contains two red points in its interior.
Similarly we can construct example of 2k-fan based on the regular (2k − 1)-gon.
π
Example 3.2. The 2k-fan consists of one angle equal to π and 2k − 1 angles equal to 2k−1
. We
start from a regular (2k − 1)-gon and distribute the red measure µ1 in its midpoints of sides
for even k (see Figure 2), or in the vertices for odd k (see Figure 3).
Let the blue measure µ2 have k − 1 points in the “inner” region of the same (2k − 1)-gon and
additional k points in the “outer” angles of the same (2k − 1)-gon that lie below the “bottom”
line. This is shown in Figures 2 and 3.
To explain this counterexample, we start with the top π angle. Any of its translations
touching more than a half of the blue measure must contain the “bottom” line of the (2k − 1)gon and therefore already contains at least k red points in the interior. So this angle cannot
make an equipartition.
4
ALEXEY BALITSKIY, ALEXEY GARBER, AND ROMAN KARASEV
Figure 1. Two measures and a 4-fan without equipartitioning.
Now consider the angle F1 next to the top angle. If, after a translation, it touches more than a
half of the red points then it must contain in the interior the k −1 “inner” blue points (it is seen
in Figures 2 and 3 for both cases) and one more blue point to the left in the corresponding outer
angle of the (2k − 1)-gon. Because if an “inner” blue point is outside of it or on its boundary
then the extension of one of its sides, passing almost through the center of the (2k − 1)-gon,
completely separates this angle from some k of red points.
As for the next angle F2 , after a translation it must touch one of the “inner” blue points and
at least one of the “outer” blue points (if it only touches “outer” blue points then it already
contains everything). Then it is seen from the picture that this angle will also contain in its
interior some k red vertices of the (2k − 1)-gon, because one of its sides is completely outside
the red polygon, and the other cuts the bigger part of it. The argument for the remaining
angles Fj is the same depending on the parity of j.
F1
F2
Figure 2. Two measures and a 4m-fan without equipartitioning.
F2k−1
ANOTHER HAM SANDWICH IN THE PLANE
F1
5
F2k−1
F2
Figure 3. Two measures and a (4m + 2)-fan without equipartitioning.
Example 3.3. Now we construct two measures for an arbitrary 4-fan without two opposite rays
that that will show that the conclusion of Theorem 2.4 fails for such fans (see Figure 4). It is
easy to establish by a small case analysis that, for any such fan, it is possible to find angles F1
and F3 out of F1 , F2 , F3 , F4 so that −F1 is contained in F3 and F3 \ (−F1 ) has two connected
components.
Now we split every measure into 3 equal parts and do the following. One red point is put to
the origin, two blue points are put in the respective components of F3 \ (−F1 ) near the origin,
two red points are put in the respective components of (−F3 ) \ F1 much farther from the origin
so that one of them is in F2 and the other is in F4 , and we put the remaining blue point in F1
very far from the origin.
Then we notice the following. If F3 , after a translation, touches two red points then it also
contains the two blue points closer to the origin. If a translated F1 touches two red points then
it definitely contains in the interior one of the closer blue points and the farthest blue point.
As for a translation of F2 (and analogously F4 ), if it touches some two of the blue points then
it contains in the interior the red point at the origin and the other red point form the same
side as F2 as well.
4. Measures with non-compact support
Using the same approach as before we can extend Theorems 1.1 and 1.2 to the case of pairs
of measures with non-compact support:
Theorem 4.1. Theorem 1.1 holds for measures with non-compact support if no line parallel to
any ray from the fan F divide both measures µ1 and µ2 into equal halves.
Proof. As in the proof of Theorem 1.1, construct the sets Sji of all possible translations of the
angle Fj that divides µi into equal parts. If µi has non-compact support, this set does not
necessarily contain two rays parallel to the sides of the angle −Fj , but it can be approximated
6
ALEXEY BALITSKIY, ALEXEY GARBER, AND ROMAN KARASEV
F3
F4
F2
F1
Figure 4. Two measures and an arbitrary 4-fan without equipartitioning.
by such rays when we go sufficiently far from the origin. Thus, as before we can construct
similar angles αji with sides on lines parallel to rays of the fan F that divides measures into
equal parts.
Assume that for some j the angle αj1 is not a subset of the angle αj2 and αj2 is not a subset of αj1 .
These angles does not have a common boundary ray, and we can find a point of intersection Sj1
and Sj2 using the same argument of odd index of intersection of a pair of curves. This completes
the proof.
Another possible argument to prove this extension is to use going to the limit argument. The
only case when we cannot choose a limit of a sequence of angles Ak is when the intersection
of Ak with every disk Dr tends to a half-plane. But in this case we would obtain a half-plane
cutting precisely half of every measure, which is assumed to be impossible.
Adding the same restriction we can modify the Theorem 1.2 for measures with non-compact
support. However, for arbitrary measures it is not always possible to find a translation of an
angle that divides both measures into equal halves. We show this below for the case of two
normally distributed measures.
Lemma 4.2. No angle with angular measure less than π can simultaneously cut a half of both
2
2
x2 +y 2
1
1
measures µ1 and µ2 with densities ρ1 (x, y) = 2π
exp(− x +y
)
and
ρ
(x,
y)
=
exp(−
).
2
2
8π
8
Proof. Assume F is a translation of the initial angle that has vertex in the origin (0, 0). Denote
as F+ (a, b) the translation of the angle F by the vector (a, b) and by fi (a, b) denote the ith
measure of the angle F+ (a, b), i.e.
ZZ
fi (a, b) :=
ρi (x, y) dxdy.
F+ (a,b)
We will use the following properties of functions f1 and f2 :
(1) f1 (a, b) = f2 (2a, 2b).
(2) If for some i fi (a, b) = 1/2, then (a, b) lies strictly inside the angle −F , centrally
symmetric to F with respect to the origin.
The first property can be easily obtained by substituting x = 2x′ and y = 2y ′ in the double
integral formula for f2 (2a, 2b). Assume that the second property fails, then there is a half-plane
π containing the angle F+ (a, b) and not containing the origin in its interior. So, µi (F+ (a, b)) <
µi (π) ≤ 1/2, which is impossible.
ANOTHER HAM SANDWICH IN THE PLANE
7
Now, assume there is a vector (a, b) such that f1 (a, b) = f2 (a, b) = 1/2. Then f2 (a, b) =
f2 (2a, 2b) = 1/2, but due to the second property, the angle F+ (2a, 2b) contains the angle
F+ (a, b) in its interior and equality f2 (a, b) = f2 (2a, 2b) is impossible.
From the above proof it is clear that we could use two arbitrary normally distributed measures
with common mean vector and proportional, but distinct, covariance matrices.
Corollary 4.3. If a fan F does not contain angle equal to π then no translation of any angle
from F can divide both measures µ1 and µ2 from the previous lemma into equal parts.
References
[1] H. Steinhaus. Sur la division des ensembles de l’espace par les plans et des ensembles plans par les cercles.
Fundam. Math, 33 (1945), 245–263.
[2] A. H. Stone and J. W. Tukey. Generalized “sandwich” theorems. Duke Mathematical Journal, 9:2 (1942),
356–359.
[3] R. Živaljević. Illumination complexes, ∆-zonotopes, and the polyhedral curtain theorem. Arxiv preprint
arXiv:1307.5138, 2013.
Dept. of Mathematics, Moscow Institute of Physics and Technology, Institutskiy per. 9,
Dolgoprudny, Russia 141700
Institute for Information Transmission Problems RAS, Bolshoy Karetny per. 19, Moscow,
Russia 127994
E-mail address: alexey [email protected]
Faculty of Mechanics and Mathematics, Moscow State University, Moscow, Leninskie gory,
1, Russia, 119991
B.N. Delone International Laboratory “Discrete and Computational Geometry”, Yaroslavl’
State University, Sovetskaya st. 14, Yaroslavl’, Russia 150000
E-mail address: [email protected]
Dept. of Mathematics, Moscow Institute of Physics and Technology, Institutskiy per. 9,
Dolgoprudny, Russia 141700
Institute for Information Transmission Problems RAS, Bolshoy Karetny per. 19, Moscow,
Russia 127994
B.N. Delone International Laboratory “Discrete and Computational Geometry”, Yaroslavl’
State University, Sovetskaya st. 14, Yaroslavl’, Russia 150000
E-mail address: r n [email protected]
URL: http://www.rkarasev.ru/en/